Finding models that realize few types

The following is a(nother) problem from an old qualifying exam in logic:

Let $$T$$ be a first-order theory in a countable language $$\mathcal{L}$$ admitting an infinite model. Show that for every cardinal $$\kappa \geq \aleph_0$$ there is a model $$\mathcal{N} \models T$$ of cardinality $$\kappa$$ such that, for every $$A \subseteq N$$, there are at most $$\vert A \vert + \aleph_0$$ types from $$S^{\mathcal{N}}_1(A)$$ realized in $$\mathcal{N}$$.

Here $$S^{\mathcal{N}}_1(A)$$ denotes the set of all complete $$1$$-types over $$A$$ in $$\text{Th}(\mathcal{N})$$ (so, a set $$p$$ of $$\mathcal{L}_A$$-formulas in one free variable belongs to $$S^{\mathcal{N}}_1(A)$$ if and only if $$p \cup \text{Th}_A(\mathcal{N})$$ is satisfiable and, for all $$\mathcal{L}_A$$-formulas $$\phi$$ in one free variable, either $$\phi \in p$$ or $$\lnot \phi \in p$$; this is a paraphrase of Marker's Definition 4.1.1).

My first instinct was to try, for each $$\kappa \geq \aleph_0$$, to find a model that is as "unsaturated" as possible. This led me to consider atomic models; however, I don't know of any existence theorems for uncountable atomic models that don't depend on specific assumptions about $$T$$. Further, because $$T$$ is not even assumed to be complete, I am doubtful whether this line of thinking is useful, since we usually don't talk about atomic or saturated models of non-complete theories.

Since the only other potentially relevant theorem I could think of was the omitting types theorem (and its generalization to higher cardinalities -- the theorem called the $$\alpha$$-omitting types theorem by Chang and Keisler), I wondered if it might be possible to use this instead; perhaps we could ensure that, in some model of the right size, many types are omitted. However, the only omitting types theorems I know assume $$A = \emptyset$$.

Is either one of these two approaches useful? If not, what would be a hint in the right direction?

• What is $S_1^\mathcal{N}(A)$? Sep 8 '20 at 3:53
• @R.Burton Thank you for your comment -- I have updated the question to explain the notation. Sep 8 '20 at 4:06